Question

given in the table are the bmi statistics for random samples of men and women. assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. complete parts (a) and (b) below. use a 0.01 significance level for both parts.
a. test the claim that males and females have the same mean body mass index (bmi).
what are the null and alternative hypotheses?
a. ( h_{0}: mu_{1} geq mu_{2} )
( h_{1}: mu_{1}<mu_{2} )
b. ( h_{0}: mu_{1}=mu_{2} )
( h_{1}: mu_{1}>mu_{2} )
c. ( h_{0}: mu_{1}
eq mu_{2} )
( h_{1}: mu_{1}<mu_{2} )
d. ( h_{0}: mu_{1}=mu_{2} )
( h_{1}: mu_{1}
eq mu_{2} )
the test statistic, ( t ), is ( square ). (round to two decimal places as needed.)

Explanation:

Step1: Calculate the test statistic formula

The formula for the test statistic \(t\) for two - sample \(t\) - test (unequal variances) is \(t=\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\)

Here, \(\bar{x}_{1} = 27.0039\), \(s_{1}=7.101911\), \(n_{1} = 40\), \(\bar{x}_{2}=24.1173\), \(s_{2}=4.141761\), \(n_{2}=40\)

Step2: Substitute the values into the formula

First, calculate the numerator: \(\bar{x}_{1}-\bar{x}_{2}=27.0039 - 24.1173=2.8866\)

Then, calculate the denominator:
\(\frac{s_{1}^{2}}{n_{1}}=\frac{7.101911^{2}}{40}=\frac{50.4471}{40}=1.261178\)
\(\frac{s_{2}^{2}}{n_{2}}=\frac{4.141761^{2}}{40}=\frac{17.1541}{40}=0.428853\)
\(\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=\sqrt{1.261178 + 0.428853}=\sqrt{1.690031}\approx1.30\)

Finally, \(t=\frac{2.8866}{1.30}\approx2.22\)

Answer:

The test statistic \(t\) is approximately \(2.22\)